5) Assume a pair of alleles (Aa) control a trait in the population. Restriction

Question

5) Assume a pair of alleles (Aa) control a trait in the population. Restriction Fragment Length Polymorphism (RFLP) DNA analyses were conducted from the blood sample provided by 1000 individuals. Alleles were identified based of size differences (Allele ‘’A’’ is 500 bp DNA fragment and allele ‘’a’’ is 400 bp with 100 bp deletion as compared ti A). Do the following observed genotype frequencies indicate that the population is Hardy-Weinberg equilibrium for this trait? Test the hypothesis using the chi-square test.
Observed Genotypes: AA (homozygous dominant )= 500 individuals, Aa (heterozygote) =250 individuals, aa (homozygous recessive ) = 250 individuals.

Explanation / Answer

Hardy–Weinberg equilibrium: A better, but equivalent, probabilistic description for the HWP is that the alleles for the next generation for any given individual are chosen randomly and independent of each other. Consider two alleles, A and a, with frequencies p and q, respectively, in the population. The different ways to form new genotypes can be derived using a Punnett square, where the fraction in each is equal to the product of the row and column probabilities. In the simplest case of a single locus with two alleles: the dominant allele is denoted A and the recessive a and their frequencies are denoted by p and q; freq(A) = p; freq(a) = q; p + q = 1. If the population is in equilibrium, then we will have freq(AA) = p2 for the AA homozygotes in the population, freq(aa) = q2 for the aa homozygotes, and freq(Aa) = 2pq for the heterozygotes. The formula is sometimes written as (p2) + (2pq) + (q2) = 1, representing the fact that probabilities (normalised frequencies for a theoretically infinite population size) must add up to one. The final three possible genotypic frequencies in the offspring become: * f(AA) = p2, * f(Aa) = 2pq, * f(aa) = q2, These frequencies are called Hardy–Weinberg frequencies (or Hardy–Weinberg proportions). This is achieved in one generation, and only requires the assumption of random mating with an infinite population size. Sometimes, a population is created by bringing together males and females with different allele frequencies. In this case, the assumption of a single population is violated until after the first generation, so the first generation will not have Hardy–Weinberg equilibrium. Successive generations will have Hardy–Weinberg equilibrium. Solution: A= 500 a= 400 Aa= 100 Observed Genotypes: AA (homozygous dominant )= 500 individuals, Aa (heterozygote) =250 individuals, aa (homozygous recessive ) = 250 individuals.

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